What is the axis of symmetry and vertex for the graph y = 2x^2 - 8x + 4y=2x2−8x+4?

1 Answer
Jess
Jan 12, 2018

Complete the square (or use (-b)/(2a)−b2a)

Explanation:

To complete the square for y=2x^2-8x+4y=2x2−8x+4:
First take out the 2 for the first two terms
y=2(x^2-4x)+4y=2(x2−4x)+4
Then take the value for b (which is 4 here), divide by 2 and write it like this:
y=2(x^2-4x+2^2-2^2)+4y=2(x2−4x+22−22)+4
They both cancel out each other so adding these two terms to the equation isn't a problem.
Within your new equation take the first term and third term (x^2x2 and 2) inside the brackets and put the sign of the second term (-−) between these two so it looks something like this:
y=2((x-2)^2-2^2)+4y=2((x−2)2−22)+4
Then simplify:
y=2(x-2)^2-4y=2(x−2)2−4

The x coordinate of the vertex is found by taking the expression within the brackets and simply doing:
0=x-20=x−2
so
x=2x=2
and the y coordinate is the number behind the brackets.
y=-4y=−4

So the coordinates of the vertex becomes:
(2, -4)(2,−4)
And the axis of symmetry:
x=2x=2

Another way to get the same answer is to use (-b)/(2a)−b2a
x=(-b)/(2a)x=−b2a
x=8/(2(2))x=82(2)
x=2x=2

and substitute 2 in into y=2x^2-8x+4y=2x2−8x+4 to find yy.

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